A triangle has corners at $(8, 5)$ $(8, 5 )$ , ( 2, -7) $, and$ $, and$ ( 1, -2 )#. If the triangle is reflected across the x-axis, what will its new centroid be?

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A triangle has corners at $(8, 5)$ $(8, 5 )$ , ( 2, -7) $, and$ $, and$ ( 1, -2 )#. If the triangle is reflected across the x-axis, what will its new centroid be?

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Answer 1 · 2018-05-18T12:49:33

$(\frac{11}{3}, \frac{4}{3})$ $(11/3,4/3)$

Explanation:

$begin by calculating the coordinates of the centroid$ $"begin by calculating the coordinates of the centroid"$

$given the coordinates of the vertices of a triangle say$ $"given the coordinates of the vertices of a triangle say"$

$(x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3})$ $(x_1,y_1),(x_2,y_2)" and "(x_3,y_3)$

$then the centroid is the average of the x and y$ $"then the centroid is the average of the x and y"$
$coordinates of the vertices$ $"coordinates of the vertices"$

$centroid = [\frac{1}{3} (x_{1} + x_{2} + x_{3}), \frac{1}{3} (y_{1} + y_{2} + y_{3})]$ $"centroid "=[1/3(x_1+x_2+x_3),1/3(y_1+y_2+y_3)]$

$\Rightarrow [\frac{1}{3} (8 + 2 + 1), \frac{1}{3} (5 - 7 - 2)] = (\frac{11}{3}, - \frac{4}{3})$ $rArr[1/3(8+2+1),1/3(5-7-2)]=(11/3,-4/3)$

$under a reflection in the x-axis$ $"under a reflection in the x-axis"$

$∙ a point (x, y) \to (x, - y)$ $• " a point "(x,y)to(x,-y)$

$\Rightarrow (\frac{11}{3}, - \frac{4}{3}) \to (\frac{11}{3}, \frac{4}{3}) \leftarrow new centroid$ $rArr(11/3,-4/3)to(11/3,4/3)larrcolor(red)"new centroid"$

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A triangle has corners at $(8, 5)$ $(8, 5 )$ , ( 2, -7) $, and$ $, and$ ( 1, -2 )#. If the triangle is reflected across the x-axis, what will its new centroid be?

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

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A triangle has corners at (8,5)(8, 5 ), ( 2, -7),and, and ( 1, -2 )#. If the triangle is reflected across the x-axis, what will its new centroid be?

1 Answer

Explanation:

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A triangle has corners at $(8, 5)$ $(8, 5 )$ , ( 2, -7) $, and$ $, and$ ( 1, -2 )#. If the triangle is reflected across the x-axis, what will its new centroid be?